A projectile is fired at a target at point C, which lies 100 feet from

Question

BHBH.jpg A projectile is fired at a target at point C, which lies 100 feet from the base of the platform from which the projectile was fired at point A. If AB = 90 feet, by what distance, in feet, did the projectile fall short of its target? A. 100 - 40 \sqrt{5} B. 100 - 40 \sqrt{3} C. 20 D. 25 5. 40

bratbg · Answer

Using Phythagoras therom

Imo - A) 100-40√5

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Mohammadmo · Answer

According to Pythagoras theorem, 
90^2=10^2-X^2
X^2=8000
X=√8000=40√5
So, BC=100-40√5
Option A

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amarsinha · Answer

100 - (\sqrt{90^2-10^2}
= 100 -\sqrt{8000}
= 100-40\sqrt{5}

ScottTargetTestPrep · Answer

We can let x = the distance from point B to the base of the platform, so by the Pythagorean theorem, we have:

x^2 + 10^2 = 90^2

x^2 + 100 = 8100

x^2 = 8000

x = √8000 = √1600 x √5 = 40√5

Since 100 is the distance from point C to the base of the platform, the projectile fell short by 100 - 40√5 feet.

Answer: A

A projectile is fired at a target at point C, which lies 100 feet from

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A projectile is fired at a target at point C, which lies 100 feet from

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